College Physics

whereΔωis thechange in angular velocityandΔtis the change in time. The units of angular acceleration are(rad/s)/s, orrad/s^2. Ifω

increases, thenαis positive. Ifωdecreases, thenαis negative.

Example 10.1 Calculating the Angular Acceleration and Deceleration of a Bike Wheel

Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a)

Calculate the angular acceleration inrad/s^2. (b) If she now slams on the brakes, causing an angular acceleration of – 87.3 rad/s^2 , how long

does it take the wheel to stop?

Strategy for (a)

The angular acceleration can be found directly from its definition inα=Δω

Δt

because the final angular velocity and time are given. We see that

Δωis 250 rpm andΔtis 5.00 s.

Solution for (a)

Entering known information into the definition of angular acceleration, we get

(10.5)

α = Δω

Δt

=

250 rpm

5.00 s

.

BecauseΔωis in revolutions per minute (rpm) and we want the standard units ofrad/s^2 for angular acceleration, we need to convertΔω

from rpm to rad/s:

(10.6)

Δω = 250rev

min

⋅2π radrev ⋅1 min

60 sec

= 26.2rads.

Entering this quantity into the expression forα, we get

α = Δω (10.7)

Δt

= 26.2 rad/s

5.00 s

= 5.24 rad/s^2.

Strategy for (b)

In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular

acceleration and solving forΔt, yielding

(10.8)

Δt=Δαω.

Solution for (b)

Here the angular velocity decreases from26.2 rad/s(250 rpm) to zero, so thatΔωis – 26.2 rad/s, andαis given to be – 87.3 rad/s^2.

Thus,

(10.9)

Δt = – 26.2 rad/s

– 87.3 rad/s^2

= 0.300 s.

Discussion

Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When

she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are

analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall—the velocity change

is large in a short time interval.

If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then
come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear
and angular acceleration are related. In circular motion, linear acceleration istangentto the circle at the point of interest, as seen inFigure 10.4.

Thus, linear acceleration is calledtangential accelerationat.

CHAPTER 10 | ROTATIONAL MOTION AND ANGULAR MOMENTUM 321